What is the general solution for sin(2θ)-cos(θ)-2sin(θ)+1=0 ?

The general solution for sin(2θ)-cos(θ)-2sin(θ)+1=0 is θ=2pin,θ= pi/6+2pin,θ=(5pi)/6+2pin

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sin(2θ)-cos(θ)-2sin(θ)+1=0

sin (2 θ) - cos (θ) - 2 sin (θ) + 1 = 0

θ = 2 πn, θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Solution steps

sin (2 θ) - cos (θ) - 2 sin (θ) + 1 = 0

Rewrite using trig identities

1 - cos (θ) - 2 sin (θ) + 2 cos (θ) sin (θ) = 0

Factor

1 - cos (θ) - 2 sin (θ) + 2 cos (θ) sin (θ) : (cos (θ) - 1) (2 sin (θ) - 1)

(cos (θ) - 1) (2 sin (θ) - 1) = 0

Solving each part separately

cos (θ) - 1 = 0 or 2 sin (θ) - 1 = 0

cos (θ) - 1 = 0 : θ = 2 πn

2 sin (θ) - 1 = 0 : θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Combine all the solutions

θ = 2 πn, θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

2 cos (x) + 1 = 0 - 2, π \leq x \leq 2 π

cos^{3} (x) = - cos (x)

csc (2 x) = 2 sin (2 x), 0 \leq x \leq 2 π

cos (x) = - \frac{12}{13}

x, z = arcsin (\frac{x}{y})

What is the general solution for sin(2θ)-cos(θ)-2sin(θ)+1=0 ?
The general solution for sin(2θ)-cos(θ)-2sin(θ)+1=0 is θ=2pin,θ= pi/6+2pin,θ=(5pi)/6+2pin